A345185 Numbers that are the sum of five third powers in nine or more ways.

5860, 6112, 6138, 6462, 6497, 6588, 6651, 6859, 6947, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8056, 8163, 8289, 8352, 8371, 8443, 8506, 8560, 8569, 8630, 8632, 8758, 8928, 8991, 9017, 9045, 9080, 9099

Offset: 1

Author

David Consiglio, Jr., Jun 10 2021

Keywords

• nonn

Examples

			6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3  = 1^3 + 3^3 + 7^3 + 12^3 + 14^3  = 1^3 + 6^3 + 6^3 + 7^3 + 16^3  = 2^3 + 2^3 + 9^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 11^3 + 15^3  = 2^3 + 8^3 + 9^3 + 9^3 + 14^3  = 3^3 + 3^3 + 3^3 + 4^3 + 17^3  = 3^3 + 5^3 + 8^3 + 11^3 + 14^3  = 8^3 + 8^3 + 8^3 + 11^3 + 12^3.
		

Links

• David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A341891, A344802, A345146, A345183, A345186, A345187, A345518.

Programs

Python

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 9])
for x in range(len(rets)):
    print(rets[x])